Sections 1 to 7: definition and existence, transform table, inverse transforms, IVPs, step functions and shifting, the Dirac delta, convolution第 1 至 7 节：定义与存在性、变换表、逆变换、初值问题、阶跃函数与平移、狄拉克 delta 函数、卷积CALC IV
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PART I · CORE TECHNIQUES第一部分 · 核心技巧Computational fluency · 28 marks计算熟练度 · 28 分
Computing and Inverting Transforms计算变换与求逆变换
Show all working. For inverse transforms, write partial fractions or completing-the-square steps explicitly before quoting the table. State any shift theorem you use.展示完整解题过程。求逆变换时，在查表之前须明确写出部分分式分解或配方步骤，并说明所用的平移定理。
Q1MEDIUMCOREdefinition integral and linearity定义积分与线性性[8 marks]
Use the definition $\mathcal{L}\{f\}(s)=\displaystyle\int_{0}^{\infty}e^{-st}f(t)\,dt$ to compute each transform directly from the integral. Do not quote table entries; derive them.利用定义 $\mathcal{L}\{f\}(s)=\displaystyle\int_{0}^{\infty}e^{-st}f(t)\,dt$ 直接从积分计算每个变换，不得直接引用变换表，须从头推导。
(a) $\mathcal{L}\{e^{3t}\}$, valid for $s>3$.在 $s>3$ 时成立。[3]
(b) $\mathcal{L}\{\cos 2t\}$, valid for $s>0$. (Use integration by parts twice, or Euler's formula.)在 $s>0$ 时成立。（用两次分部积分，或利用欧拉公式。）[3]
(c)Hence, using linearity, find $\mathcal{L}\{5e^{3t}-3\cos 2t\}$.由此，利用线性性，求 $\mathcal{L}\{5e^{3t}-3\cos 2t\}$。[2]
Q2MEDIUMCOREfirst shift theorem and transform table第一平移定理与变换表[8 marks]
Apply the first shift theorem $\mathcal{L}\{e^{at}f(t)\}(s)=F(s-a)$ together with standard table entries. State clearly which entry you start from and what shift you apply.利用第一平移定理 $\mathcal{L}\{e^{at}f(t)\}(s)=F(s-a)$ 结合标准变换表计算。须明确说明所用的表项及所施加的平移量。
Q3HARDCOREinverse transform: partial fractions and completing the square逆变换：部分分式与配方[6 marks]
Find $f(t)=\mathcal{L}^{-1}\{F(s)\}$ for each $F(s)$. Perform partial fractions or complete the square before inverting. Show all algebra.对每个 $F(s)$，求 $f(t)=\mathcal{L}^{-1}\{F(s)\}$。在求逆变换之前须进行部分分式分解或配方，展示完整的代数过程。
(a) $F(s)=\dfrac{3s+1}{s^{2}-s-2}$ [3]
(b) $F(s)=\dfrac{2s+6}{s^{2}+4s+13}$ [3]
Q4HARDCOREsecond shift theorem and unit step第二平移定理与单位阶跃函数[6 marks]
The unit step function is $u_{c}(t)=0$ for $t<c$ and $1$ for $t\ge c$. The second shift theorem states $\mathcal{L}\{u_{c}(t)f(t-c)\}=e^{-cs}F(s)$.单位阶跃函数定义为：$t<c$ 时 $u_{c}(t)=0$，$t\ge c$ 时 $u_{c}(t)=1$。第二平移定理为 $\mathcal{L}\{u_{c}(t)f(t-c)\}=e^{-cs}F(s)$。
(c)Write $g(t)=u_{1}(t)\cdot t^{2}$ in the form required by the second shift theorem and find $\mathcal{L}\{g\}$. (You must rewrite $t^{2}$ as a polynomial in $t-1$.)将 $g(t)=u_{1}(t)\cdot t^{2}$ 改写为第二平移定理所需的形式，并求 $\mathcal{L}\{g\}$。（须将 $t^{2}$ 改写为关于 $t-1$ 的多项式。）[2]
PART II · DEFINITIONS AND PROOF第二部分 · 定义与证明Rigorous arguments · 26 marks严密论证 · 26 分
Derivations and Theoretical Results推导与理论结果
These items are graded on the logic of the argument. Quote every formula you apply. Integration by parts must state $u$ and $dv$ explicitly. In proofs, verify boundary terms and convergence conditions.本部分按论证逻辑评分。须引用所用的每个公式。分部积分必须明确写出 $u$ 和 $dv$。在证明中须验证边界项和收敛条件。
Q5HARDPROOFderivative rule via integration by parts通过分部积分推导导数法则[8 marks]
Throughout, assume $y(t)$ is of exponential order and piecewise continuous on $[0,\infty)$, so that $e^{-st}y(t)\to 0$ as $t\to\infty$ for sufficiently large $s$. Let $Y(s)=\mathcal{L}\{y\}(s)$.全题假设 $y(t)$ 在 $[0,\infty)$ 上为指数阶且分段连续，从而对足够大的 $s$，有 $e^{-st}y(t)\to 0$（$t\to\infty$）。令 $Y(s)=\mathcal{L}\{y\}(s)$。
(a)Starting from the definition $\mathcal{L}\{y'\}(s)=\displaystyle\int_{0}^{\infty}e^{-st}y'(t)\,dt$, integrate by parts to derive the formula $\mathcal{L}\{y'\}=sY(s)-y(0)$. State your choice of $u$ and $dv$, evaluate the boundary term, and justify why the boundary term at $t\to\infty$ vanishes.从定义 $\mathcal{L}\{y'\}(s)=\displaystyle\int_{0}^{\infty}e^{-st}y'(t)\,dt$ 出发，用分部积分推导公式 $\mathcal{L}\{y'\}=sY(s)-y(0)$。须说明 $u$ 和 $dv$ 的选取，计算边界项，并说明 $t\to\infty$ 处边界项为零的原因。[4]
(b)Apply the result of (a) to $y''=(y')'$ to derive $\mathcal{L}\{y''\}=s^{2}Y(s)-sy(0)-y'(0)$. Show each substitution step.将 (a) 的结论应用于 $y''=(y')'$，推导 $\mathcal{L}\{y''\}=s^{2}Y(s)-sy(0)-y'(0)$，展示每个代入步骤。[4]
Q6HARDPROOFfirst shift theorem: proof and application第一平移定理：证明与应用[10 marks]
The first shift theorem (s-shift) states: if $\mathcal{L}\{f(t)\}=F(s)$ for $s>s_0$, then $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$ for $s>s_0+a$.第一平移定理（$s$ 平移）：若 $\mathcal{L}\{f(t)\}=F(s)$ 在 $s>s_0$ 时成立，则 $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$ 在 $s>s_0+a$ 时成立。
(a)Prove the first shift theorem directly from the definition by substituting $e^{at}f(t)$ into $\int_{0}^{\infty}e^{-st}(\cdot)\,dt$ and collecting exponentials.从定义出发直接证明第一平移定理：将 $e^{at}f(t)$ 代入 $\int_{0}^{\infty}e^{-st}(\cdot)\,dt$ 并合并指数项。[3]
(b)Prove the second shift theorem: $\mathcal{L}\{u_{c}(t)f(t-c)\}=e^{-cs}F(s)$ for $c\ge 0$. Start from the definition, change the variable of integration to $\tau=t-c$, and justify the new lower limit.证明第二平移定理：$\mathcal{L}\{u_{c}(t)f(t-c)\}=e^{-cs}F(s)$，$c\ge 0$。从定义出发，令 $\tau=t-c$ 换元，并说明新下限的合理性。[4]
(c)Use one of the two theorems you just proved to find $\mathcal{L}\{e^{-t}\cos 2t\}$ without quoting the table directly. Show which theorem applies and carry through the algebra.利用刚证明的两个定理之一，不直接引用变换表，求 $\mathcal{L}\{e^{-t}\cos 2t\}$。说明适用哪个定理并完成代数运算。[3]
Q7HARDPROOFconvolution theorem and integral identity卷积定理与积分恒等式[8 marks]
The convolution of $f$ and $g$ is $(f*g)(t)=\displaystyle\int_{0}^{t}f(\tau)g(t-\tau)\,d\tau$. The convolution theorem states $\mathcal{L}\{f*g\}=F(s)\cdot G(s)$.$f$ 与 $g$ 的卷积为 $(f*g)(t)=\displaystyle\int_{0}^{t}f(\tau)g(t-\tau)\,d\tau$。卷积定理为 $\mathcal{L}\{f*g\}=F(s)\cdot G(s)$。
(a)Evaluate $(1*e^{t})(t)=\displaystyle\int_{0}^{t}e^{\tau}\,d\tau$ directly, and verify the convolution theorem by showing that $\mathcal{L}\{(1*e^{t})(t)\}=\mathcal{L}\{1\}\cdot\mathcal{L}\{e^{t}\}$.直接计算 $(1*e^{t})(t)=\displaystyle\int_{0}^{t}e^{\tau}\,d\tau$，并通过验证 $\mathcal{L}\{(1*e^{t})(t)\}=\mathcal{L}\{1\}\cdot\mathcal{L}\{e^{t}\}$ 来确认卷积定理。[4]
(b)Use the convolution theorem to evaluate $\mathcal{L}^{-1}\!\left\{\dfrac{1}{s(s^{2}+4)}\right\}$ without partial fractions: identify $F(s)$ and $G(s)$, state $f$ and $g$, and compute the convolution integral.利用卷积定理，不用部分分式，计算 $\mathcal{L}^{-1}\!\left\{\dfrac{1}{s(s^{2}+4)}\right\}$：确定 $F(s)$ 和 $G(s)$，写出 $f$ 和 $g$，并计算卷积积分。[4]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合题 · 28 分
IVPs, Impulses, and Step-Function Forcing初值问题、冲击与阶跃函数强迫
Set up each problem by taking the Laplace transform of the entire equation, apply initial conditions immediately, and solve the algebraic equation for $Y(s)$ before inverting. Verify initial conditions in your final answer.每题须对整个方程取拉普拉斯变换建立方程，立即代入初始条件，解出 $Y(s)$ 的代数方程后再求逆变换。在最终答案中验证初始条件。
Q8HARDAPPLIEDsolving a second-order IVP by Laplace transform用拉普拉斯变换求解二阶初值问题[8 marks]
Use the Laplace transform to solve the initial value problem
$$ y'' - 3y' + 2y = 4e^{t}, \qquad y(0)=1,\quad y'(0)=0. $$用拉普拉斯变换求解初值问题
$$ y'' - 3y' + 2y = 4e^{t}, \qquad y(0)=1,\quad y'(0)=0. $$
(a)Take the Laplace transform of both sides, apply initial conditions, and obtain an expression for $Y(s)=\mathcal{L}\{y\}$.对两侧取拉普拉斯变换，代入初始条件，得到 $Y(s)=\mathcal{L}\{y\}$ 的表达式。[3]
(b)Decompose $Y(s)$ by partial fractions.对 $Y(s)$ 进行部分分式分解。[3]
(c)Invert each term to obtain $y(t)$, and verify that $y(0)=1$ and $y'(0)=0$.逐项求逆变换得到 $y(t)$，并验证 $y(0)=1$ 及 $y'(0)=0$。[2]
Q9HARDAPPLIEDIVP with step-function forcing阶跃函数强迫下的初值问题[10 marks]
Solve the initial value problem
$$ y'' + 4y = g(t), \qquad y(0)=0,\quad y'(0)=0, $$
where the forcing switches on at $t=\pi$:
$$ g(t)=\begin{cases}0, & 0\le t<\pi,\\ 1, & t\ge\pi.\end{cases} $$求解初值问题
$$ y'' + 4y = g(t), \qquad y(0)=0,\quad y'(0)=0, $$
其中强迫项在 $t=\pi$ 处接入：
$$ g(t)=\begin{cases}0, & 0\le t<\pi,\\ 1, & t\ge\pi.\end{cases} $$
(a)Write $g(t)$ in terms of the unit step function $u_{\pi}(t)$ and find $\mathcal{L}\{g\}$.用单位阶跃函数 $u_{\pi}(t)$ 表示 $g(t)$，并求 $\mathcal{L}\{g\}$。[2]
(b)Take the Laplace transform of the ODE, apply initial conditions, and solve for $Y(s)$. Decompose by partial fractions.对常微分方程取拉普拉斯变换，代入初始条件，解出 $Y(s)$，并进行部分分式分解。[4]
(c)Use the inverse second shift theorem to find $y(t)$. Write your answer as a piecewise function and verify $y(0)=0$ and $y'(0)=0$.利用第二平移逆定理求 $y(t)$，将答案写成分段函数，并验证 $y(0)=0$ 及 $y'(0)=0$。[4]
Q10HARDAPPLIEDimpulse response with the Dirac delta狄拉克 delta 函数的冲击响应[10 marks]
A damped oscillator is struck by an impulsive force at time $t=2$. Solve
$$ y'' + 2y' + 5y = \delta(t-2), \qquad y(0)=0,\quad y'(0)=0. $$
Recall that $\mathcal{L}\{\delta(t-a)\}=e^{-as}$ for $a\ge 0$.一阻尼振子在 $t=2$ 时受到冲击力的作用，求解
$$ y'' + 2y' + 5y = \delta(t-2), \qquad y(0)=0,\quad y'(0)=0. $$
已知 $\mathcal{L}\{\delta(t-a)\}=e^{-as}$（$a\ge 0$）。
(a)Transform the equation, apply initial conditions, and solve for $Y(s)$.对方程取变换，代入初始条件，解出 $Y(s)$。[3]
(b)Factor the denominator polynomial and complete the square. Identify the relevant table entry for the inverse transform.对分母多项式因式分解并配方，确定逆变换所用的相关变换表项。[3]
(c)Apply the inverse second shift theorem to find $y(t)$. Express the answer in terms of $u_{2}(t)$ and verify that $y(t)=0$ for $0\le t<2$.利用第二平移逆定理求 $y(t)$，用 $u_{2}(t)$ 表示答案，并验证 $0\le t<2$ 时 $y(t)=0$。[4]